PS 9.1 | Electric Current and Ohm’s Law

Student Solution Key

Solutions

Conceptual Explanation:

Electric current ($I$): rate of charge flow through a conductor.
Voltage ($V$): electric potential difference or “pressure” driving the current.
Resistance ($R$): opposition to current flow in the conductor.
Units: $I$ (A = C/s), $V$ (V = J/C), $R$ ($\Omega$ = V/A)

a) Calculate the resistance in the circuit.

$V = IR$ (Ohm’s Law)

$V = 230~\text{V}$
$I = 1.75~\text{A}$

$V = IR$ (rearrange to solve for $R$)
$R = \frac{V}{I}$
$R = \frac{230~\text{V}}{1.75~\text{A}}$
$R = 131.428…~\Omega$
Answer: $R = 131~\Omega$

3
($230$ has 3 sig figs)

b) Explain why doubling the voltage would double the current.

Ohm’s Law: $I = \frac{V}{R}$. If $R$ is constant, doubling $V$ doubles $I$.

a) What is the current flowing through the circuit?

$V = IR$ (Ohm’s Law)

$V = 9.00~\text{V}$
$R_1 = 470~\Omega$

$V = IR$ (rearrange to solve for $I$)
$I = \frac{V}{R}$
$I = \frac{9.00~\text{V}}{470~\Omega}$
$I = 0.01914…~\text{A}$
Answer: $I = 0.0191~\text{A}$

3
(Limited by $9.00$ V)

b) If the resistor is replaced with a $1.00~\text{k}\Omega$ resistor, how does the current change?

$V = IR$ (Ohm’s Law)

$V = 9.00~\text{V}$
$R_2 = 1.00~\text{k}\Omega = 1000~\Omega$

$V = IR$ (rearrange to solve for $I$)
$I = \frac{V}{R}$
$I = \frac{9.00~\text{V}}{1000~\Omega}$
$I = 0.00900~\text{A}$
Answer: $I = 0.00900~\text{A}$
Current decreases as resistance increases.

3
(Limited by $9.00$ V)

What is the resistance of a microwave drawing $12.5~\text{A}$ at $120~\text{V}$?

$V = IR$ (Ohm’s Law)

$V = 120~\text{V}$
$I = 12.5~\text{A}$

$V = IR$ (rearrange to solve for $R$)
$R = \frac{V}{I}$
$R = \frac{120~\text{V}}{12.5~\text{A}}$
$R = 9.600~\Omega$
Answer: $R = 9.60~\Omega$

3
(Limited by $12.5$ A)
Six
a) What voltage must be applied across a $33.0~\Omega$ LED requiring $15.0~\text{mA}$?

$V = IR$ (Ohm’s Law)

$R = 33.0~\Omega$
$I = 15.0~\text{mA} = 0.0150~\text{A}$

$V = IR$
$V = (0.0150)(33.0)$
$V = 0.495~\text{V}$
Answer: $V = 0.495~\text{V}$

3
(Matches input values)

b) Why is the direction of current flow important for LEDs?

LEDs only allow current flow in one direction due to their semiconductor properties. Reverse current can damage the device.

A battery supplies $24.0~\text{V}$ to a circuit with two identical resistors. If the total current is $0.800~\text{A}$, what is the resistance of each resistor?

$R_\text{total} = \frac{V}{I}$

$V = 24.0~\text{V}$
$I = 0.800~\text{A}$

$R_\text{total} = \frac{24.0~\text{V}}{0.800~\text{A}} = 30.0~\Omega$
Answer: $R_\text{total} = 30.0~\Omega$

 3<br>
(Matches given values)

$R_\text{series} = \frac{R_\text{total}}{2}$

$R_\text{total} = 30.0~\Omega$

$R_\text{each} = \frac{30.0~\Omega}{2} = 15.0~\Omega$
Answer: $R_\text{each} = 15.0~\Omega$

$R_\text{parallel} = 2 \times R_\text{total}$

$R_\text{total} = 30.0~\Omega$

$R_\text{each} = 2 \times 30.0~\Omega = 60.0~\Omega$
Answer: $R_\text{each} = 60.0~\Omega$

a) Calculate the resistance of a car starter motor ($12.6~\text{V}$, $185~\text{A}$).

$V = IR$ (Ohm’s Law)

$V = 12.6~\text{V}$
$I = 185~\text{A}$

$V = IR$ (rearrange to solve for $R$)
$R = \frac{V}{I}$
$R = \frac{12.6~\text{V}}{185~\text{A}}$
$R = 0.0681…\Omega$
Answer: $R = 0.0681~\Omega$

3
(Matches given values)

b) How much charge flows through it in $3.00~\text{s}$?

$Q = I \times t$

$I = 185~\text{A}$
$t = 3.00~\text{s}$

$Q = It$
$Q = (185)(3.00)$
$Q = 555~\text{C}$

3
(Limited by $3.00$ s)

a) What is the effective resistance of a laptop charging circuit ($19.5~\text{V}$, $3.33~\text{A}$)?

$R = \frac{V}{I}$

$V = 19.5~\text{V}$
$I = 3.33~\text{A}$

$R = \frac{V}{I}$
$R = \frac{19.5~\text{V}}{3.33~\text{A}}$
$R = 5.8558…\Omega$
Answer: $R = 5.86~\Omega$

3
(Matches input values)

b) How much charge is transferred in $2.00~\text{hours}$?

$Q = I \times t$

$I = 3.33~\text{A}$
$t = 2.00~\text{hr} = 7200~\text{s}$

$Q = It$
$Q = (3.33)(7200)$
$Q = 23976~\text{C}$
Answer: $Q = 2.40\times10^{4}~\text{C}$

3
(Limited by $2.00$ hr)